the density of f is 1 on (0,1)
E(U^{-t})=\int_0^1 u^{-t} f(u)du where f(u)=1
then just use calc 1 \int_0^1 u^{-t} du =u^{-t+1}/(-t+1)
Hi,
I'm just need to work out a step taken in the answer to the following question:
A random variable U has the uniform distribution on the range (0, 1). Find the moment
generating function of W, where W is given by W = − ln U, and hence identify the
distribution of W.
Hint: The moment generating function, with argument t, of the exponential distribution with mean θ is (1 − θt)^(−1), for t < θ^(−1)
Now the answer goes like this:
Moment Generating Function = E(exp(tW))
=E(exp(t(-logU)))
=E((exp(logU))^(-t))
=E(U^(-t))
Now from here i don't fully understand how they get to the next step:
=int(1*u^(-t),u=0..1)
and from here they go on to show
Moment Generating Function = 1/(1-t)
I understand how to find expectancies, by integrating, but i am unsure why it is 1*, if someone could explain this to me i would be very greatful.
p.s sorry for not using Latex but it doesn't seem to be working.